Question

Using the quadratic formula to solve x2 = 5 – x, what are the values of x?

Answer 1

Hello : 
the discriminat of each quadratic equation : ax²+bx+c=0 ....(a ≠ 0) is :
Δ = b² -4ac
1 )  Δ > 0  the equation has two reals solutions : x =  (-b±√Δ)/2a
2 ) Δ = 0 : one solution : x = -b/2a
3 ) Δ < 0 : no reals solutions
in this exercice : x² = 5-x
x² +x-5 = 0    .... a =1   b= 1   c=-5
calculate Δ..................................

Answer 2

Answer:

$x_{1}=\frac{-1б 4.58}{2} = 3.58\\\\x_{2}=\frac{-1б -4.58}{2} =-5.58$

Step-by-step explanation:

Hello

In elementary algebra, the quadratic formula is the solution of the quadratic equation.

let a polynom

$ax^{2} +bx+c=0$

this can be solved d by using the quadratic equation formula

$x= \frac{-bб \sqrt{b^{2}-4ac } }{2a}$

Let

$x^{2} =5-x$

Step 1

do the equation=0

$x^{2} =5-x\\x^{2}+x-5=0$

define

$a=1\\b=1\\c=-5$

Sep two

put the values into the equation

$x= \frac{-bб \sqrt{b^{2}-4ac } }{2a}\\\\\\x= \frac{-(1)б \sqrt{(1)^{2}-4(1)(-5) } }{2*1}\\x= \frac{-1б \sqrt{1+20 } }{2}\\x= \frac{-1б 4.58}{2}\\\\x_{1}=\frac{-1б 4.58}{2} = 3.58\\\\x_{2}=\frac{-1б -4.58}{2} =-5.58$

Have a good day.